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# prove tan x cos 2x + tan x = sin 2 x

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prove tan x cos 2x + tan x + sin 2 x

edited by nlbradshaw0430  Apr 13, 2017
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I think you're missing something, here......there's nothing to prove....

CPhill  Apr 13, 2017
edited by CPhill  Apr 13, 2017
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Maybe it should just be simplify????

tan x cos 2x + tan x + sin 2 x

$$tan x *cos 2x + tan x + sin 2 x\\ =\frac{sinx}{cosx} *(cos^2x-sin^2x)+ \frac{sinx}{cosx} + 2sinxcosx\\ =sinxcosx-\frac{sin^3x}{cosx}+ \frac{sinx}{cosx} + 2sinxcosx\\ =3sinxcosx-\frac{sin^3x}{cosx}+ \frac{sinx}{cosx} \\ =3sinxcosx- \frac{sinx}{cosx} (sin^2x-1)\\ =3sinxcosx- \frac{sinx}{cosx} (-cos^2x)\\ =3sinxcosx+ (sinxcosx)\\ =4sinxcosx\\ =2sin2x$$

Lets see if I could have done this more simply....

$$tan x cos 2x + tan x + sin 2 x\\ = tan x (cos^2x-sin^2x) + tan x + sin 2 x\\ = tan x (cos^2x-sin^2x+1) + sin 2 x\\ = tan x (cos^2x+cos^2x) + sin 2 x\\ =\frac{sinx}{cosx} (2cos^2x) + sin 2 x\\ =2sinxcosx + sin 2 x\\ =sin2x + sin 2 x\\ =2sin2x \\$$

So it can be seen that

tan x cos 2x + tan x = sin 2 x

which I guess was meant to be the question in the first place.   (as hectictar and CPhill have already noted.  Thanks guys. :)

Melody  Apr 13, 2017
edited by Melody  Apr 13, 2017
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tan x cos 2x + tan x + sin 2 x      ......simplified  ???

[sinx/ cosx] [ cos^2x - sin^2x]  + sinx/cosx +  2sinx cosx

sinxcosx  - sin^3x/ cosx   +  sinx/cosx + 2sinxcosx

3sinxcosx  -  [sinx ( sin^2x - 1)] / cosx

3sinxcosx  -  [ -sinx (1 - sin^2x) ]/ cosx

3sinxcosx  + [sinx cos^2x]/cosx

3sinxcosx + sinxcosx  =

4sinxcosx       or       2sin(2x)

CPhill  Apr 13, 2017
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OK...I see now....

tan x cos 2x + tan x = sin 2 x

tanx (cos 2x + 1)

tan x ( 2cos^2x - 1 + 1)

tanx ( 2cos^2x)

2 * [sinx / cosx] * cos^2x

2sin x cos x    =    sin 2x

Thanks to hectictar for bringing this to my attention....!!!!

CPhill  Apr 13, 2017
edited by CPhill  Apr 13, 2017